If \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{b}}\) are two unit vectors such that \(\overline{\mathrm{a}}+2 \overline{\mathrm{~b}}\) and \(5 \bar{a}-4 \bar{b}\) are perpendicular to each other, then the angle between \(\bar{a}\) and \(\bar{b}\) is

Let \(\theta\) be the angle between \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{b}}\).
Since \(\overline{\mathrm{c}}=\overline{\mathrm{a}}+2 \overline{\mathrm{~b}}\) and \(\overline{\mathrm{d}}=5 \overline{\mathrm{a}}-4 \overline{\mathrm{~b}}\) are perpendicular to each other.
\(\therefore \overline{\mathrm{c}} \cdot \overline{\mathrm{~d}}=0 \)
\( \Rightarrow(\overline{\mathrm{a}}+2 \overline{\mathrm{~b}}) \cdot(5 \overline{\mathrm{a}}-4 \overline{\mathrm{~b}})=0 \)
\( \Rightarrow 5(\overline{\mathrm{a}} \cdot \overline{\mathrm{a}})+6(\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}})-8(\overline{\mathrm{~b}} \cdot \overline{\mathrm{~b}})=0 \)
\( \Rightarrow 5|\overline{\mathrm{a}}|^2+6|\overline{\mathrm{a}}||\overline{\mathrm{b}}| \cos \theta-8|\overline{\mathrm{~b}}|^2=0 \)
\( \Rightarrow 5+6 \cos \theta-8=0 \)
\( \Rightarrow \cos \theta=\frac{1}{2} \)
\( \Rightarrow \theta=\frac{\pi}{3}\)